POJ 3026 Borg Maze BFS+MST

Description

The Borg is an immensely powerful race of enhanced humanoids from the delta quadrant of the galaxy. The Borg collective is the term used to describe the group consciousness of the Borg civilization. Each Borg individual is linked to the collective by a sophisticated subspace network that insures each member is given constant supervision and guidance. 

Your task is to help the Borg (yes, really) by developing a program which helps the Borg to estimate the minimal cost of scanning a maze for the assimilation of aliens hiding in the maze, by moving in north, west, east, and south steps. The tricky thing is that the beginning of the search is conducted by a large group of over 100 individuals. Whenever an alien is assimilated, or at the beginning of the search, the group may split in two or more groups (but their consciousness is still collective.). The cost of searching a maze is definied as the total distance covered by all the groups involved in the search together. That is, if the original group walks five steps, then splits into two groups each walking three steps, the total distance is 11=5+3+3.

Input

On the first line of input there is one integer, N <= 50, giving the number of test cases in the input. Each test case starts with a line containg two integers x, y such that 1 <= x,y <= 50. After this, y lines follow, each which x characters. For each character, a space `` '' stands for an open space, a hash mark ``#'' stands for an obstructing wall, the capital letter ``A'' stand for an alien, and the capital letter ``S'' stands for the start of the search. The perimeter of the maze is always closed, i.e., there is no way to get out from the coordinate of the ``S''. At most 100 aliens are present in the maze, and everyone is reachable.

Output

For every test case, output one line containing the minimal cost of a succesful search of the maze leaving no aliens alive.

Sample Input

26 5##### #A#A### # A##S  ####### 7 7#####  #AAA####    A## S ####     ##AAA########  

Sample Output

811

这题A的太困难啦。。昨天开始做，今天早上才A掉。。。

题意：给出一个图，S代表Borg，去抓A。抓到的A就变成Borg（也可以去抓A），然后只有A和S是在一起的时候才可以分开。问最少多少步可以把A全部抓到。

思路：先用最短路求出每两个点之间的距离，用一个dist[x][y]数组保存。x,y分别是该点的序号。

然后再用prime求出最小生成树，即全部抓到最少的步数。因为是求MST，所以A和S是一样的。所以就一同编号了。

#include <iostream>#include <cstdio>#include <algorithm>#include <string>#include <cmath>#include <cstring>#include <queue>#include <set>#include <vector>#include <stack>#include <map>#include <iomanip>#define PI acos(-1.0)#define Max 55#define inf 1<<28using namespace std;struct kdq{    int x,y;} ;int point[Max][Max],num,numx,numy;char Map[Max][Max],first[Max*Max];bool visit[Max][Max],v[Max*Max];int move[Max*Max][Max*Max],n,m,dis[Max*Max],dist[Max*Max][Max*Max],movex[4]= {1,-1,0,0},movey[4]= {0,0,1,-1};queue<kdq>q;int inmap(int i,int j){    if(i>=1&&j>=0&&i<m&&j<n)        return 1;    return 0;}void init(){    int i,j;    for(i=0; i<num+10; i++)    {        for(j=0; j<num+10; j++)            move[i][j]=0,visit[i][j]=0;    }}void bfs(int x,int y){    kdq date;    date.x=x,date.y=y;    init();    visit[x][y]=1;    q.push(date);    while(!q.empty())    {        kdq temp=q.front();        q.pop();        for(int   i=0; i<4; i++)        {            int tx=temp.x+movex[i];            int ty=temp.y+movey[i];            if(inmap(tx,ty)&&Map[tx][ty]!='#'&&!visit[tx][ty])            {                move[tx][ty]=move[temp.x][temp.y]+1;                visit[tx][ty]=1;                if(point[tx][ty])//如果该点是A或者S。即point[temp.x][y=temp.y]!=0。则更新点(x,y)到点（temp.x,temp.y）的距离。                {                    dist[point[x][y]][point[tx][ty]]=move[tx][ty];                }                kdq xx;                xx.x=tx;                xx.y=ty;                q.push(xx);            }        }    }}int prime(int numx,int numy,int num)//以numx,numy为起点，其实任意点都可以。{    int i,j;    memset(v,0,sizeof(v));    for(i=1; i<=num; i++)        dis[i]=dist[point[numx][numy]][i];    v[point[numx][numy]]=1;    int ans=0;    for(i=0; i<num-1; i++)    {        int index=-1;        int min=inf;        for(j=1; j<=num; j++)        {            if(min>dis[j]&&!v[j])                min=dis[j],index=j;        }        if(index==-1)            return ans;        v[index]=1;        ans+=min;        for(j=1; j<=num; j++)        {            if(dis[j]>dist[index][j]&&!v[j])                dis[j]=dist[index][j];        }    }    return ans;}void read(){    int i,j;    cin>>n>>m;    num=0;    gets(first);//万恶的空格。。。    memset(point,0,sizeof(point));    for(i=0; i<m; i++)    {        gets(Map[i]);    }}void solve(){    int i,j;    for(i=0; i<m; i++)        for(j=0; j<n; j++)        {            if(Map[i][j]=='A'||Map[i][j]=='S')                point[i][j]=++num,numx=i,numy=j;//cout<<point[i][j]<<endl;//对每个点A或者S进行编号。        }    for(i=0; i<m; i++)    {        for(j=0; j<n; j++)        {            if(point[i][j])                bfs(i,j);//求点（i,j）到其他所有点的最短路径。        }    }    cout<<prime(numx,numy,num)<<endl;}int main(){    int i,j,k,l,T;    cin>>T;    while(T--)    {        read();        solve();    }    return 0;}



这道题做完的感受就是，有些细节的地方太粗心了。比如初始化的时候，一开始point[][]没有memset。而且代码写的太凌乱了。这道题最难的地方应该就是给任意两个点求出最短路然后更新。
不过，说实话。。刚才A掉这道题的时候还是很兴奋的。。毕竟A了整整一天半~~~~
	
				
	
					   
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